Does this sequence converge to $0$?

Question

In a similar vain as my previous question, I was wondering whether the following converges to $0:$

$$ a_n = \left\{ \begin{array}{l l} 1 & \quad n=2^k\,\text{for some}\,k\in\mathbb{N}_0\\ 0 & \quad \text{otherwise} \end{array} \right.$$

First terms are:

$$1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,1,0\cdots$$

In other words, the $1$'s get more and more separated. Again, as in the other question, intuitively it would seem that "at infinity", the string of $0$'s is infinite, but since intuition can often be erroneous in cases like these, I am not convinced. Can someone enlighten me here?

Follow-up:

Can I conclude from the commments and answers that a sequence of integers that converges to $0$ must be $0$ for $n$ sufficiently large?

How can it possibly converge to zero? No matter how far out you go, there are ones and zeros. If it converged, all the values would end up 'close' to each other. — copper.hat, Mar 07 '13 at 19:33
@copper.hat Sorry, I know it's maybe a stupid question but I wasn't sure. I am having trouble with the concept of a limit I think. — Little miss sunshine, Mar 07 '13 at 19:35
Yes to your follow-up. You need to work on your intuition regarding limits. — copper.hat, Mar 07 '13 at 19:38
Yes to the followup. Use the $\varepsilon-N$ definition of limit with $\varepsilon=1$ to show that there must be an $N$ for each $|a_n|<1$ for all $n>N$. Since the $a_i$ are integers, that means $a_n=0$ for $n>N$. — Thomas Andrews, Mar 07 '13 at 19:40

score 2 · Accepted Answer · answered Mar 07 '13 at 19:35

For $\epsilon=\frac12$, there is no $N\in \mathbb N$ with the property that for all indices $n>N$ we have $|a_n|<\epsilon$. Even though they are somewhat rare, terms $a_n$ with $a_n=1>\epsilon$ occur again and again. (To be explicit, we can consider $n=2^N$, which is $>N$ and gives us $a_n=1$).

There is something behind this: The sequence indeed converges to zero in a less strict sense so to speak, see Cesáro mean.

score 2 · Answer 2 · answered Mar 07 '13 at 19:48

The other answers are good, this is more of a side-topic or another place to look.

There is a notion of something called the "Cesàro mean." Given a convergent sequence, the Cesàro mean is just the limit. But the Cesàro mean also gives a value for some non-convergent sequences.

For example, $-1,1,-1,1,\dots$ has Cesàro mean zero, but does not have a limit.

Your series has Cesàro mean zero, too.

score 1 · Answer 3 · answered Mar 07 '13 at 19:34

No. A sequence converges to zero if and only if we can eventually make all the terms which are large enough as close to zero as we want. In symbols, $a_n \to a$ if for all $\epsilon > 0$, there exists a constant $M$ so that $n > M$ implies $|a_n - a| < \epsilon$.

This cannot happen with your example, since every entry that is a power of two will be $1$.

Does this sequence converge to $0$?

3 Answers3